If Delta x is the uncertainty in position and | vedclass

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(C) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$.Given that the uncertainty in position $\Delta x$ i

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$\frac{1}{2} \sqrt{\frac{m h}{\pi}}$

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(C) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$.Given that the uncertainty in position $\Delta x$ is equal to the uncertainty in velocity $\Delta v$,so $\Delta x = \Delta v$.Since $\Delta p = m \Delta v$,we can substitute $\Delta v = \Delta x$ into the uncertainty relation:$\Delta x \cdot (m \Delta x) = \frac{h}{4 \pi}$$m (\Delta x)^2 = \frac{h}{4 \pi}$$(\Delta x)^2 = \frac{h}{4 \pi m}$$\Delta x = \frac{1}{2} \sqrt{\frac{h}{\pi m}}$Since $\Delta p = m \Delta v$ and $\Delta v = \Delta x$,then $\Delta p = m \Delta x$.Substituting the value of $\Delta x$:$\Delta p = m \cdot \frac{1}{2} \sqrt{\frac{h}{\pi m}}$$\Delta p = \frac{1}{2} \sqrt{\frac{m^2 h}{\pi m}}$$\Delta p = \frac{1}{2} \sqrt{\frac{m h}{\pi}}$

If $\Delta x$ is the uncertainty in position and $\Delta v$ is the uncertainty in velocity of a particle are equal,the correct expression for uncertainty in momentum for the same particle is

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Why was a change in the Bohr Model of atom required? Due to which important development$(s)$,was the concept of movement of an electron in an orbit replaced by the concept of probability of finding an electron in an orbital? What is the name given to the changed model of atom?

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Reason : The path of an electron in an atom is clearly defined.

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The minimum uncertainty in the speed of an electron in a one-dimensional region of length $2 a_{0}$ (where $a_{0} = \text{Bohr radius} = 52.9 \ pm$) is $km \ s^{-1}$. (Given: Mass of electron $= 9.1 \times 10^{-31} \ kg$,Planck's constant $h = 6.63 \times 10^{-34} \ J \ s$)

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